# How do you find the exact values of cos(5pi/12) using the half angle formula?

Jul 23, 2015

$\cos \left(\frac{5 \pi}{12}\right) = \frac{\sqrt{2 - \sqrt{3}}}{2}$

#### Explanation:

By the half angle formula:
$\textcolor{w h i t e}{\text{XXXX}}$$\cos \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \left(\theta\right)}{2}}$

If $\frac{\theta}{2} = \frac{5 \pi}{12}$
$\textcolor{w h i t e}{\text{XXXX}}$then $\theta = \frac{5 \pi}{6}$

Note that $\frac{5 \pi}{6}$ is a standard angle in quadrant 2 with a reference angle of $\frac{\pi}{6}$
so $\cos \left(\frac{5 \pi}{6}\right) = - \cos \left(\frac{\pi}{6}\right) = - \frac{\sqrt{3}}{2}$

Therefore
$\textcolor{w h i t e}{\text{XXXX}} \cos \left(\frac{5 \pi}{12}\right) = \pm \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}$

$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} = \pm \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}}$

$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} = \pm \sqrt{\frac{2 - \sqrt{3}}{4}}$

$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} = \pm \frac{\sqrt{2 - \sqrt{3}}}{2}$

Since $\frac{5 \pi}{12} < \frac{\pi}{2}$
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{5 \pi}{12}$ is in quadrant 1
$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow \cos \left(\frac{5 \pi}{12}\right)$ is positive
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$(the negative solution is extraneous)

May 12, 2018

$\cos \left(\frac{5 \pi}{12}\right) = \frac{1}{2} \sqrt{2 - \sqrt{3}} = \frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right)$

#### Explanation:

$\frac{5 \pi}{12} \setminus \times \frac{{360}^{\circ}}{2 \pi} = {75}^{\circ} = {150}^{\circ} / 2$

How? First I sigh and point out that this is yet another trig problem that assumes the student knows only the two cliche triangles, 30/60/90 and 45/45/90.

You, dear reader, might object that this question only requires us to know one triangle, 30/60/90, which includes ${150}^{\circ}$ in the second quadrant. And you'd be correct; to answer the question as asked we use the half angle formula:

$\cos \left(\frac{a}{2}\right) = \pm \sqrt{\frac{1}{2} \left(1 + \cos a\right)}$

$\cos {75}^{\circ} = \pm \sqrt{\frac{1}{2} \left(1 + \cos {150}^{\circ}\right)}$

$= \pm \sqrt{\frac{1}{2} \left(1 - \cos {30}^{\circ}\right)} = \pm \sqrt{\frac{1}{2} \left(1 - \frac{\sqrt{3}}{2}\right)}$

$= \pm \frac{1}{2} \sqrt{2 - \sqrt{3}}$

There are a couple of things less than ideal here. First we have the ambiguous $\pm$ which we have to disambiguate by saying ${75}^{\circ}$ is in the first quadrant, so must have a positive cosine.

$\cos {75}^{\circ} = \frac{1}{2} \sqrt{2 - \sqrt{3}}$

Where does the minus sign come from? If instead of ${150}^{\circ}$ we started with the coterminal $- {210}^{\circ}$ the half angle of $- {105}^{\circ}$ has the negative cosine.

The second issue is the nested square root. There's a theorem that says $\sqrt{a + b \sqrt{d}}$ can be denested if ${a}^{2} - {b}^{2} d$ is a perfect square, which it is here. But rather than getting distracted by denesting, which is fun but not germane, let's talk about the proper way to do this problem.

We were asked to use the half angle formula. It would be better if we had used the sum angle formula. This gives an unambiguous result with no nested square root:

$\cos {75}^{\circ} = \cos \left({30}^{\circ} + {45}^{\circ}\right)$

$= \cos {30}^{\circ} \cos {45}^{\circ} - \sin {30}^{\circ} \sin {45}^{\circ}$

$= \left(\setminus \frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{2}}{2}\right) - \left(\frac{1}{2}\right) \frac{\sqrt{2}}{2}$

$= \frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right)$

There's a bit of a mystery about how Ramanujan derived some of his radical expressions. Perhaps we get a hint in our first Ramanujan-like result:

$\frac{1}{2} \sqrt{2 - \sqrt{3}} = \frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right)$